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Lomka1309
@Lomka1309
July 2022
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4 в степени x+2 в степени x-20=0
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hlopushinairina
4^x+2^x-20=0;⇒(2²)^x-2^x-20=0;
2^x=y;⇒y²-y-20=0;
y₁₂=1/2⁺₋√[(1+80)/4]=1/2⁺₋√81/4=1/2⁺₋9/2;
y₁=10/2=5;⇒2^x=5;⇒x=log₂5;
y₂=1/2-9/2=-8/2=-4;⇒2^x=-4;⇒x=log₂(-4);⇒loga b⇒b>0⇒2^x≠-4;
ответ:x=log₂5
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Answers & Comments
2^x=y;⇒y²-y-20=0;
y₁₂=1/2⁺₋√[(1+80)/4]=1/2⁺₋√81/4=1/2⁺₋9/2;
y₁=10/2=5;⇒2^x=5;⇒x=log₂5;
y₂=1/2-9/2=-8/2=-4;⇒2^x=-4;⇒x=log₂(-4);⇒loga b⇒b>0⇒2^x≠-4;
ответ:x=log₂5